Simplify and expand the following expression: $ \dfrac{4}{y - 10}+ \dfrac{4}{3y - 9}- \dfrac{4y}{y^2 - 13y + 30} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{4}{3y - 9} = \dfrac{4}{3(y - 3)}$ We can factor the quadratic in the third term: $ \dfrac{4y}{y^2 - 13y + 30} = \dfrac{4y}{(y - 10)(y - 3)}$ Now we have: $ \dfrac{4}{y - 10}+ \dfrac{4}{3(y - 3)}- \dfrac{4y}{(y - 10)(y - 3)} $ The least common multiple of the denominators is: $ (y - 10)(y - 3)$ In order to get the first term over $(y - 10)(y - 3)$ , multiply by $\dfrac{3(y - 3)}{3(y - 3)}$ $ \dfrac{4}{y - 10} \times \dfrac{3(y - 3)}{3(y - 3)} = \dfrac{12(y - 3)}{(y - 10)(y - 3)} $ In order to get the second term over $(y - 10)(y - 3)$ , multiply by $\dfrac{y - 10}{y - 10}$ $ \dfrac{4}{3(y - 3)} \times \dfrac{y - 10}{y - 10} = \dfrac{4(y - 10)}{(y - 10)(y - 3)} $ In order to get the third term over $(y - 10)(y - 3)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{4y}{(y - 10)(y - 3)} \times \dfrac{3}{3} = \dfrac{12y}{(y - 10)(y - 3)} $ Now we have: $ \dfrac{12(y - 3)}{(y - 10)(y - 3)} + \dfrac{4(y - 10)}{(y - 10)(y - 3)} - \dfrac{12y}{(y - 10)(y - 3)} $ $ = \dfrac{ 12(y - 3) + 4(y - 10) - 12y} {(y - 10)(y - 3)} $ Expand: $ = \dfrac{12y - 36 + 4y - 40 - 12y}{3y^2 - 39y + 90} $ $ = \dfrac{4y - 76}{3y^2 - 39y + 90}$